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Thursday, May 16, 2013

Week of May 13th Math Problem: Finding the Lengths of the Major and Minor Axes for a Tricky Ellipse

For many of us, we begin our discussion of conic sections in high school with circles and ellipses. We first learn the equation for a circle is of the general form:

(x - h)2 + (y - k)2 = r2

where (h, k) is the set of coordinates for the center of the circle and r is the radius of the circle. We then move on to ellipses, and find the general equation is

[(x - h)2] / a2 + [(y - k)2] / b2 = 1

when a > b (horizontally elongated ellipse; switch the positions of the x and y terms for a vertically elongated ellipse). The length of the major axis is therefore 2a and the minor axis 2b. Here is a horizontally-elongated ellipse (image from Wikipedia):

How would one answer the following question?

What are the lengths of the major and minor axes for the following ellipse?

4x2 + 8x + 16y2 - 64y - 13 = 0

We would first complete the square by adding a total of 68 to both sides (split up as 4 and 64) to reveal two factorable trinomials:

(4x2 + 8x + 4) + (16y2 - 64y + 64) - 13 = 68

Factoring out a 4 and 16 from the first and second trinomials respectively, and adding 13 to both sides, this becomes:

4(x2 + 2x + 1) + 16(y2 - 4y + 4) = 81

Factoring the trinomials

4(x + 1)2 + 16(y -2)2 = 81

and dividing both sides by 81

(4/81)(x + 1)2 + (16/81)(y -2)2 = 1

This is nearly in the form requisite for an ellipse. How do we rewrite this to show a2and b2 in the denominators? The key is to rewrite the equations first as